Procedure to generate an inverse for a non-square matrix

Question

We know that if a matrix is $m \times n$ where $m=n$ then it has an inverse if its full rank.
But the interesting fact for me is about non-square matrices, where $m\neq n$, and also I searched on the whole web for it and found pseudo inverse and so on$\ldots$

However, I found an amazing example from the site which has a right inverse and also its transpose which has a left inverse (the transpose is that matrix right inverse also which the multiplication will result in identity matrix): $$\begin{bmatrix}1 &0\end{bmatrix}$$
My main questions is:

1. Is there any procedure to compute the left or right inverse of a matrix (same as the example) when it’s full rank in the case of rows OR columns? I mean not tje pseudo inverse but the real inverse which will result in the Identity matrix.
2. What should be the condition of that matrix in which we will compute its left OR right inverses?

Answer 1

First, let me clarify that there is no “real inverse” of an $m\times n$ matrix where $m\neq n$. The inverse is defined only for square matrices.

Let $A$ be a matrix of size $m\times n$ with $m\neq n$. If $\operatorname{rank}(A)=\min\{m,n\}$, the Moore-Penrose pseudoinverse will produce left- or right-inverses that yield the identity matrix of appropriate size, and they have a simple expression. In particular,

1. If $m>n$ and $\operatorname{rank}(A)=n$, we say $A$ is full column rank and it has a left-inverse $$A^{\dagger}=(A^*A)^{-1}A^*$$ which satisfies $A^{\dagger}A=I_n$.
2. If $m<n$ and $\operatorname{rank}(A)=m$, we say $A$ is full row rank and it has a right-inverse given by $$A^{\dagger}=A^*(AA^*)^{-1}$$ that satisfies $AA^{\dagger}=I_m$.

In your example, let $A=\begin{bmatrix}1&0\end{bmatrix}$. Its rank is clearly 1, and hence $A$ is full row rank. Then, its right inverse is given by $$A^{\dagger}=\begin{bmatrix}1&0\end{bmatrix}^* \Big(\begin{bmatrix}1&0\end{bmatrix}\begin{bmatrix}1&0\end{bmatrix}^*\Big)^{-1}=\begin{bmatrix}1\\0\end{bmatrix}\left(\begin{bmatrix}1&0\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix}\right)^{-1}=\begin{bmatrix}1\\0\end{bmatrix}(1)^{-1}=\begin{bmatrix}1\\0\end{bmatrix}\,.$$

If we consider its transpose, i.e., $B=\begin{bmatrix}1\\0\end{bmatrix}$, then $B$ is full column rank and has left inverse given by

$$B^{\dagger}=\left(\begin{bmatrix}1&0\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix}\right)^{-1}\begin{bmatrix}1\\0\end{bmatrix}^*=(1)^{-1}\begin{bmatrix}1&0\end{bmatrix}=\begin{bmatrix}1&0\end{bmatrix}\,.$$